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Principle deep dive

Modulus / absolute value behaviour

Piecewise splitting at zero. The principle behind the 2023 modulus spike — broadest cross-chapter reach in the bank (10 chapters tagged), plus invocations across App of Derivatives, Apps of Integration, Functions, Linear Inequalities, Probability, Quad Eq and Sets & Relations.

questions in the bank
106
tagged HARD
14%
chapter spread
10
worked examples below
4

When to reach for it

The expression contains |x|, [x], or a function defined piecewise at zero or an integer.

Why this principle matters

|x| splits at zero: equals x for x ≥ 0, equals −x for x < 0. That single piecewise split drives every modulus question in NDA. Left and right limits diverge at the split point; the function is continuous there but not differentiable.

The principle has been on a tear since 2023 — modulus jumped from 4 q/paper-set to 15. It now appears across 8 chapters: Limits, Continuity, Differentiation, Definite Integration, Apps of Integration, Differential Equations, Functions, and Sets. The technique is the same everywhere: split at the discontinuity, handle each piece separately, recombine.

The greatest-integer function [x] is the modulus's discrete cousin. It's also piecewise — constant on each [n, n+1) interval, jumping by 1 at each integer. Most NDA traps around [x] live at the integer boundaries (limits, derivatives, integrals).

4 worked examples from the bank

Each example demonstrates the principle on a real past-year question. Click to reveal the answer, then the solution.

Example 1Limits & ContinuityEASY
Which one of the following is correct regarding lim⁡x→3∣x−3∣x−3\displaystyle\lim_{x\to3}\dfrac{|x-3|}{x-3}?

[Q74 · Sep · 2024]

Example 2DifferentiationEASY
If f(x)=e∣x∣f(x)=e^{|x|}, then which one of the following is correct?

[Q84 · Apr · 2021]

Example 3Limits & ContinuityMODERATE
If f(x)=x2+x+∣x∣xf(x) = \frac{x^2+x+|x|}{x}, then what is lim⁡x→0f(x)\lim_{x\to 0} f(x) equal to?

[Q86 · Sep · 2022]

Example 4Limits & ContinuityHARD
Let f(x)=∣x∣+1f(x)=|x|+1, g(x)=[x]−1g(x)=[x]-1, h(x)=f(x)⋅g(x)h(x)=f(x)\cdot g(x). What is lim⁡x→0−h(x)+lim⁡x→0+h(x)\displaystyle\lim_{x\to0^-}h(x)+\lim_{x\to0^+}h(x) equal to?

[Q88 · Apr · 2024]

Variants to recognise

Same principle, different surfaces. Pattern-match these on test day.

  • Piecewise definition of |x|

    |x| = x for x ≥ 0, −x for x < 0. The split point matters; everything else is algebra.

  • Left vs right limit at the split

    lim x→0⁻ |x|/x = −1; lim x→0⁺ |x|/x = +1. Two-sided limit doesn't exist. NDA exploits this.

  • |x| is continuous, not differentiable at 0

    The graph has a corner. Left derivative = −1, right derivative = +1, so f' undefined at x = 0.

  • Greatest integer [x]

    [x] = n for x ∈ [n, n+1). Discontinuous at integers, with jump size 1. {x} = x − [x] is the fractional part.

Drill every modulus / absolute value behaviour question

106 questions from the bank — paginated, with cart and Word-export support.

Drill the 106 questions

Related principles

Often combined with this one — drill these next if you found the examples above tractable.